Chapter 17 | 871Analysis We denote the entrance, throat, and exit states by 1, t, and 2,
respectively, as shown in Fig. 17–60.
(a) Since the inlet velocity is negligible, the inlet stagnation and static states
are identical. The ratio of the exit-to-inlet stagnation pressure isIt is much smaller than the critical-pressure ratio, which is taken to be
P*/P 01 0.546 since the steam is superheated at the nozzle inlet. There-
fore, the flow surely is supersonic at the exit. Then the velocity at the throat
is the sonic velocity, and the throat pressure isAt the inlet,Also, at the throat,Then the throat velocity is determined from Eq. 17–3 to beThe flow area at the throat is determined from the mass flow rate relation:At state 2s,The enthalpy of the steam at the actual exit state is (see Chap. 7)Therefore,Then the exit velocity and the exit area becomeA 2 m#v 2
V 21 2.5 kg>s 21 0.67723 m^3 >kg 2
929.8 m>s18.21 10 ^4 m^2 18.21 cm^2V 2 221 h 01 h 22
B321 3248.42816.1 2 kJ>kg4a1000 m^2 >s^2
1 kJ>kgb929.8 m>sP 2 300 kPa
h 2 2816.1 kJ>kgf¬
v 2 0.67723 m^3 >kg
s 2 7.2019 kJ>kg#K0.933248.4h 2
3248.42783.6¡ h 2 2816.1 kJ>kghNh 01 h 2
h 01 h 2 sP 2 sP 2 300 kPa
s 2 ss 1 7.1292 kJ>kg#Kf h 2 s2783.6 kJ>kgAtm#vt
Vt1 2.5 kg/s 21 0.2420 m^3 /kg 2
585.8 m/s10.33 10 ^4 m^2 10.33 cm^2Vt 221 h 01 ht 2
B321 3248.43076.8 2 kJ/kg4a1000 m^2 >s^2
1 kJ>kgb585.8 m/sPt1.09 MPa
st7.1292 kJ>kg#Kf¬
ht3076.8 kJ>kg
vt0.24196 m^3 >kgP 1 P 01 2 MPa
T 1 T 01 400°Cf¬
h 1 h 01 3248.4 kJ>kg
s 1 sts 2 s7.1292 kJ>kg#KPt0.546P 01 1 0.546 21 2 MPa 2 1.09 MPaP 2
P 01300 kPa
2000 kPa0.15sh1P 2 = 300 kPa2
2 stPtP 1 =^P 01= 2 MPaT 1 = 400°CP 1 = 2 MPaV 1 ≅ 0
STEAM
Throatm· = 2.5 kg/shN = 93%FIGURE 17–60
Schematic and h-sdiagram for
Example 17–16.cen84959_ch17.qxd 4/21/05 11:08 AM Page 871